AMC 10 Step by Step

2011 AMC 10B Problem 4

P1-10~1 minPrint

LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip, it turned out that LeRoy had paid AA dollars and Bernardo had paid BB dollars, where A<BA < B . How many dollars must LeRoy give to Bernardo so that they share the costs equally?

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Key insight

Each person should pay half of the total. LeRoy gives Bernardo the difference between that half and what he already paid.

Solution

Try real numbers first to see what happens. Say LeRoy paid A=10A = 10 dollars and Bernardo paid B=30B = 30 dollars.

  1. The trip cost 10+30=4010 + 30 = 40 dollars, so each should pay 2020.
  2. LeRoy paid only 1010, so he owes 20−10=1020 - 10 = 10 more. He gives Bernardo 1010.
  3. Check: Bernardo paid 3030 and gets 1010 back, so he paid 2020 too.

Now do the same steps with letters.

4. The total is A+BA + B, so each person's fair share is A+B2\dfrac{A + B}{2}.
5. LeRoy has paid AA, so he still owes
A+B2−A=A+B−2A2=B−A2.\frac{A + B}{2} - A = \frac{A + B - 2A}{2} = \frac{B - A}{2}.
6. With the test numbers, this gives 30−102=10\frac{30 - 10}{2} = 10, which matches.

The answer is (C) B−A2\boxed{\textbf{(C)}\ \dfrac{B - A}{2}}.

Why this works

To make two amounts equal, you move half of the gap between them from the bigger one to the smaller one. Moving the whole gap would just swap which person is ahead. Testing easy numbers like 1010 and 3030 is a quick way to check any answer written with letters.

Alternative approach

Plug A=10A = 10 and B=30B = 30 into every choice and look for 1010. (A) gives 2020, (B) gives −10-10, (C) gives 1010, (D) gives 2020, (E) gives 4040. Only (C) matches.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The trip cost A+BA + B dollars in total, so each person's fair share is A+B2\dfrac{A + B}{2}.

LeRoy has paid only AA, so he owes
A+B2−A=B−A2.\frac{A + B}{2} - A = \frac{B - A}{2}.
This is positive because A<BA < B, and after the transfer Bernardo's net payment is B−B−A2=A+B2B - \frac{B - A}{2} = \frac{A + B}{2} as well.

The answer is (C) B−A2\boxed{\textbf{(C)}\ \dfrac{B - A}{2}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.